Initial height: 400 m
Each time it hits the ground, it rebounds 75% the distance it has fallen. Let us say this distance is d, then the new height is:
[tex]\begin{gathered} h=75\text{\% of }d=\frac{75}{100}\cdot d \\ \Rightarrow h=\frac{3}{4}d \end{gathered}[/tex]
If the initial height is 400 m, then the subsequent heights are given by the recurrence equation:
[tex]\begin{gathered} h_0=400 \\ h_n=(\frac{3}{4})^n\cdot h_0 \\ \Rightarrow h_n=400\cdot(\frac{3}{4})^n \end{gathered}[/tex]
And the total distance traveled D is:
[tex]\begin{gathered} D=h_0+2\cdot\sum ^{\infty}_{n\mathop=1}\lbrack(\frac{3}{4})^n\cdot h_0\rbrack \\ \Rightarrow D=400+800\cdot\sum ^{\infty}_{n\mathop{=}1}(\frac{3}{4})^n \end{gathered}[/tex]
Now, let us analyze the sum term:
[tex]\sum ^{\infty}_{n\mathop{=}1}(\frac{3}{4})^n=\frac{3}{4}+(\frac{3}{4}_{})^2+(\frac{3}{4})^3+\cdots_{}[/tex]
From the infinite geometric sequence:
[tex]\begin{gathered} \sum ^{\infty}_{n\mathop=0}r^n=\frac{1}{1-r} \\ \Rightarrow\sum ^{\infty}_{n\mathop{=}1}r^n=\frac{1}{1-r}-1 \end{gathered}[/tex]
Where r < 1. From our problem, r = 3/4 < 1, then:
[tex]\begin{gathered} \sum ^{\infty}_{n\mathop{=}1}(\frac{3}{4})^n=\frac{1}{1-\frac{3}{4}}-1=\frac{1}{\frac{1}{4}}-1=4-1 \\ \Rightarrow\sum ^{\infty}_{n\mathop{=}1}(\frac{3}{4})^n=3 \end{gathered}[/tex]
Finally, using this result:
[tex]D=400+800\cdot3=400+2400=2800[/tex]
